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Permutations and Combinations

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Permutations and Combinations

by divyasood11 » Sat Dec 27, 2014 3:32 am
Hi

Please help with the following problem:

In how many ways can the letters of the word ASSASSINATION be arranged
so that all the S's are together ?

My answer- 10!/3!.4!.2!.2!

OA- 10!/3!.2!.2!

My query:

Taking all 's' together we get 9 distinct spaces and one group of s. So 10 places arranged in 10! ways. Further it should be divided by the number of repetitions of A (3!), S (4!), I (2!), N(2!).

Why does the OA not account for division by 4! which represents the 4 S's.


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by MartyMurray » Sat Dec 27, 2014 5:05 am
divyasood11 wrote:Hi

Please help with the following problem:

In how many ways can the letters of the word ASSASSINATION be arranged
so that all the S's are together ?

My answer- 10!/3!.4!.2!.2!

OA- 10!/3!.2!.2!

My query:

Taking all 's' together we get 9 distinct spaces and one group of s. So 10 places arranged in 10! ways. Further it should be divided by the number of repetitions of A (3!), S (4!), I (2!), N(2!).

Why does the OA not account for division by 4! which represents the 4 S's.


Thank You
You already have the four 5's locked together as a unit. They are no longer considered four; they are considered one. Also, there are not multiple unique letters within that unit, but rather one letter that repeats four times. So there will be only one arrangement of the four letters within the unit.

With that in mind, the word ASSASSINATION has 13 letters, but your formula for finding the arrangements correctly has as the numerator 10! rather than 13! Why? To take into account the four S's' being together all the time. So the S's are already considered in the formula, as one element, rather than as four elements that can be arranged in multiple ways.

So with the four S's in a unit that has only one arrangement, there are only ten elements that can be arranged, and within that ten the A's, I's and N's repeat, giving us the formula 10!/3!2!2!.

By the way, if instead of four S's in that unit, there were four unique letters that could be arranged in multiple ways, you would still use as part of your formula 10!/3!2!2!. The difference would be that to account for the multiple arrangements of letters within the unit you would have to multiply by 4! to get 4!10!/3!2!2!.
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by GMATGuruNY » Sat Dec 27, 2014 5:20 am
divyasood11 wrote: In how many ways can the letters of the word ASSASSINATION be arranged
so that all the S's are together ?
Since the 4 S's must be in adjacent positions, glue them together as follows:
SSSS.
The effect is that the 4's become a SINGLE element in the arrangement.
Now count the number of ways to arrange the 10 elements SSSS, A, A, A, I, I, N, N, O, T.
The number of ways to arrange 10 distinct elements = 10!.
But the elements here are not all distinct.
There are 3 identical A's, 2 identical I's, and 2 identical N's.
When an arrangement includes identical elements, we must divide by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap places, the arrangement does not change, reducing the total number of unique arrangements.
Here, we must divide by 3! to account for the 3 identical A's, by 2! to account for the 2 identical I's, and by another 2! to account for the 2 identical N's:
10!/3!2!2!.
Why does the OA not account for division by 4! which represents the 4 S's
Since the 4 S's serve as a SINGLE element in the arrangement, there is no need to divide by the number of ways the 4 S's can be arranged.
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by Brent@GMATPrepNow » Sat Dec 27, 2014 9:40 am
Once we recognize that SSSS is a SINGLE ELEMENT and can be treated as 1 letter, we can apply the MISSISSIPPI, which goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

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by mbawisdom » Sun Dec 28, 2014 6:00 pm
You set the problem up correctly but forgot to that the 4 S' should be treated as 1 letter. As a result there is no need to divide by 4!.
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